Link to actual problem [1288] \[ \boxed {\left (x^{2}+4 x +3\right ) y^{\prime \prime }-\left (-x^{2}+4 x +5\right ) y^{\prime }-\left (2+x \right ) y=0} \] With initial conditions \begin {align*} [y \left (-2\right ) = 2, y^{\prime }\left (-2\right ) = -1] \end {align*}
With the expansion point for the power series method at \(x = -2\).
type detected by program
{"second order series method. Ordinary point", "second order series method. Taylor series method"}
type detected by Maple
[[_2nd_order, _with_linear_symmetries]]
Maple symgen result This shows Maple’s found \(\xi ,\eta \) and the corresponding canonical coordinates \(R,S\)\begin{align*} \\ \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{-x} \operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right ) \left (1+x \right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{x} y}{\operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right ) \left (1+x \right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{-x} \operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right ) \left (1+x \right ) \left (\int \frac {\left (x +3\right )^{8} {\mathrm e}^{x}}{\left (1+x \right )^{2} \operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right )^{2}}d x \right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{x} y}{\operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right ) \left (1+x \right ) \left (\int \frac {\left (x +3\right )^{8} {\mathrm e}^{x}}{\left (1+x \right )^{2} \operatorname {HeunC}\left (2, 1, -9, -6, -\frac {1}{2}, -\frac {x}{2}-\frac {1}{2}\right )^{2}}d x \right )}\right ] \\ \end{align*}